我在榆树中得到了一个未被捕获的类型错误，并且不知道为什么。
我从api解码json字符串; api给出了rostars的列表，每个rostar都 planningId或flexplanningId。我希望在列表上进行映射，并根据planning或id为每个planningId提供唯一的flexplanningId，具体取决于它具有的内容。这是代码：
记录定义和解码器：
type alias Rostar =
{ employee : Employee }
type alias Employee =
{ week : Week
, name : String
, id : Int
, contractHours : Float
}
type alias Week =
{ monday : List Planning
, tuesday : List Planning
, wednesday : List Planning
, thursday : List Planning
, friday : List Planning
, saturday : List Planning
, sunday : List Planning
}
type alias Planning =
{ time : String
, planningId : Maybe Int
, groupId : Int
, groupName : String
, flex : Bool
, employeeTimeslotId : Maybe Int
, flexplanningId : Maybe Int
, employeeId : Maybe Int
, id : Maybe Int
}
responseDecoder : Decoder (List Rostar)
responseDecoder =
list rostarDecoder
rostarDecoder : Decoder Rostar
rostarDecoder =
decode Rostar
|> required "employee" employeeDecoder
employeeDecoder : Decoder Employee
employeeDecoder =
decode Employee
|> required "rostar" weekDecoder
|> required "name" string
|> required "id" int
|> required "contract_hours" float
weekDecoder : Decoder Week
weekDecoder =
decode Week
|> required "monday" (list planningDecoder)
|> required "tuesday" (list planningDecoder)
|> required "wednesday" (list planningDecoder)
|> required "thursday" (list planningDecoder)
|> required "friday" (list planningDecoder)
|> required "saturday" (list planningDecoder)
|> required "sunday" (list planningDecoder)
planningDecoder : Decoder Planning
planningDecoder =
decode Planning
|> required "time" string
|> optional "planning_id" (nullable int) Nothing
|> required "group_id" int
|> required "group_name" string
|> required "flex" bool
|> optional "employee_timeslot_id" (nullable int) Nothing
|> optional "flexplanning_id" (nullable int) Nothing
|> required "employee_id" (nullable int)
|> hardcoded Nothing
映射：
update : Msg -> Model -> ( Model, Cmd Msg )
update msg model =
case msg of
HandleFeedResponse response ->
let
assignPlanningId : Planning -> Planning
assignPlanningId planning =
case planning.planningId of
Just id ->
{ planning | id = Just (id + 10000000) }
Nothing ->
case planning.flexplanningId of
Just id ->
{ planning | id = Just (id + 90000000) }
Nothing ->
{ planning | id = Nothing }
planningWithId : List Planning -> List Planning
planningWithId day =
List.map assignPlanningId day
mapWeek : Week -> Week
mapWeek week =
{ week
| monday = planningWithId week.monday
, tuesday = planningWithId week.tuesday
, wednesday = planningWithId week.wednesday
, thursday = planningWithId week.thursday
, friday = planningWithId week.friday
, saturday = planningWithId week.saturday
, sunday = planningWithId week.sunday
}
updateResponse : List Rostar
updateResponse =
List.map
(\r ->
let
employee =
{ employee | week = mapWeek employee.week }
in
{ r | employee = employee }
)
response
check =
Debug.log "updatedResponse" updateResponse
in
{ model | rostar = updateResponse } ! []
这是我得到的错误：
未捕获的TypeError：无法读取属性＆＃39;周＆＃39;未定义的 块引用
感谢您的帮助！
答案 0 :(得分：1)
我认为您的问题是由let映射函数中employee的{{1}}绑定引起的。标签updateResponse已经存在，因此该行导致递归定义。
employee在Elm 0.18中，这是一个编译错误，它会为您提供详细的错误消息，而不是为运行时错误留下可能性：
在1个模块中检测到错误。
let employee = { employee | week = mapWeek employee.week }
-- BAD RECURSION ------------------------------------------------------ Main.elm直接根据自身定义，导致无限循环。employee也许你正试图改变一个变量？榆树没有变异，所以当我 请参阅
132|> employee = 133| { employee | week = mapWeek employee.week }定义的employee，我将其视为递归 定义。尝试给新值一个新名称！也许你想要一个递归值？要定义
employee我们需要知道什么employee是，所以让我们扩展它。等等，但现在我们需要知道employee是的，所以让我们扩展它......这将继续无限！要真正了解正在发生的事情以及如何解决问题，请查看： https://github.com/elm-lang/elm-compiler/blob/0.18.0/hints/bad-recursion.md
在0.18之前，我会看到这些类型的奇数＆＃34;未定义＆＃34;意外执行某种意外递归时的运行时错误。在0.18中，他们为一些最基本的问题添加了编译器检查。
答案 1 :(得分：0)
这可能不是您错误的来源，
但是您的员工解码器说它需要一个rostar变量，其中包含week。它是否正确？或者它应该被称为week？
以下是代码段：
employeeDecoder : Decoder Employee
employeeDecoder =
decode Employee
-- |> required "rostar" weekDecoder -- is this correct?
|> required "week" weekDecoder -- what I would have expected
|> required "name" string
|> required "id" int
|> required "contract_hours" float